Problem: Vlad is playing on a swing set. His horizontal distance $D(t)$ (in $\text{m}$ ) from the center (where being behind the center means a negative distance) as a function of time $t$ (in seconds) can be modeled by a sinusoidal expression of the form $a\cdot\cos(b\cdot t)+d$. At $t=0$, when he pushes off, he is $1\text{ m}$ behind the center, which is as far back as he goes. The swing reaches the center $\dfrac{\pi}{6}$ seconds later. Find $D(t)$. $\textit{t}$ should be in radians. $D(t) = $
Explanation: The strategy First, we should convert the given information about the real-world context into mathematical terms of the sinusoidal function and its graph. Then, we should use the given information to find the amplitude, midline, and period of the function's graph. Finally, we should find $a$, $b$, and $d$ in the expression $a\cos(b\cdot t)+d$ by considering the features we found. Converting the given information into mathematical terms At $t=0$, the swing is $1\text{ m}$ behind the center. This means the graph of the function passes through $(0,-1)$. We are given that this is the farthest point behind the center, which corresponds to a minimum point of the graph. $\dfrac{\pi}{6}$ seconds later (which means $t=\dfrac{\pi}{6}$ ) the distance is $0\text{ m}$. This corresponds to the point $\left(\dfrac{\pi}{6},0\right)$. We are given that this is the middle of the swing, which corresponds to the midline of the graph. In conclusion, the graph has a minimum point at $(0,-1)$ and then intersects its midline at $\left(\dfrac{\pi}{6},0\right)$. Determining the amplitude, midline, and period The midline intersection is at $y={0}$, so this is the midline. The minimum point is $1$ units below the midline, so the amplitude is ${1}$. The minimum point is $\dfrac{\pi}{6}$ units to the left of the midline intersection, so the period is $4\cdot \dfrac{\pi}{6}={\dfrac{2\pi}{3}}$. [Why did we multiply by 4?] Determining the parameters in $a\cos(b\cdot t)+d$ Since the minimum at $t=0$ is followed by a midline intersection, we know that $a<0$. [How do we know that?] The amplitude is ${1}$, so $|a|={1}$. Since $a<0$, we can conclude that $a=-1$. The midline is $y={0}$, so $d=0$. The period is ${\dfrac{2\pi}{3}}$, so $b=\dfrac{2\pi}{\left({\dfrac{2\pi}{3}}\right)}=3$. The answer $D(t)=-\cos\left(3t\right)$